60. n 个骰子的点数【难】

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difficulty: 简单
edit_url: https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9860.%20n%E4%B8%AA%E9%AA%B0%E5%AD%90%E7%9A%84%E7%82%B9%E6%95%B0/README.md

面试题 60. n 个骰子的点数

题目描述

把n个骰子扔在地上，所有骰子朝上一面的点数之和为s。输入n，打印出s的所有可能的值出现的概率。

你需要用一个浮点数数组返回答案，其中第 i 个元素代表这 n 个骰子所能掷出的点数集合中第 i 小的那个的概率。

示例 1:

输入: 1
输出: [0.16667,0.16667,0.16667,0.16667,0.16667,0.16667]

示例 2:

输入: 2
输出: [0.02778,0.05556,0.08333,0.11111,0.13889,0.16667,0.13889,0.11111,0.08333,0.05556,0.02778]

限制：

1 <= n <= 11

解法

1.题目不太好理解，可以先看这个题：https://www.acwing.com/problem/content/76/
2.促进一点理解的题解：https://leetcode.cn/problems/nge-tou-zi-de-dian-shu-lcof/solutions/637778/jian-zhi-offer-60-n-ge-tou-zi-de-dian-sh-z36d/

方法一：动态规划

我们定义 $f[i][j]$ 表示投掷 $i$ 个骰子，点数和为 $j$ 的方案数。那么我们可以写出状态转移方程：

$$f[i][j]=\sum _{k=1}^{6}f[i-1][j-k]$$

其中 $k$ 表示当前骰子的点数，而 $f[i-1][j-k]$ 表示投掷 $i-1$ 个骰子，点数和为 $j-k$ 的方案数。

初始条件为 $f[1][j]=1$，表示投掷一个骰子，点数和为 $j$ 的方案数为 $1$。

最终，我们要求的答案即为 $\frac{f[n][j]}{6^n}$，其中 $n$ 为骰子个数，而 $j$ 的取值范围为 $[n,6n]$。

时间复杂度 $O(n^2)$，空间复杂度 $O(6n)$。其中 $n$ 为骰子个数。

Python3

class Solution:def dicesProbability(self, n: int) -> List[float]:f = [[0] * (6 * n + 1) for _ in range(n + 1)] #最大情况：投掷n个骰子，最大和为6nfor j in range(1, 7):f[1][j] = 1 #投掷1个骰子，各个和的方案数都是1for i in range(2, n + 1):for j in range(i, 6 * i + 1): #1)投掷i个筛子，和的范围为[i,6*i]for k in range(1, 7): #2）核心递推：方案数的累加if j - k >= 0:f[i][j] += f[i - 1][j - k]m = pow(6, n) #3)n个骰子的总方案数return [f[n][j] / m for j in range(n, 6 * n + 1)]

Java

class Solution {public double[] dicesProbability(int n) {int[][] f = new int[n + 1][6 * n + 1];for (int j = 1; j <= 6; ++j) {f[1][j] = 1;}for (int i = 2; i <= n; ++i) {for (int j = i; j <= 6 * i; ++j) {for (int k = 1; k <= 6; ++k) {if (j >= k) {f[i][j] += f[i - 1][j - k];}}}}double m = Math.pow(6, n);double[] ans = new double[5 * n + 1];for (int j = n; j <= 6 * n; ++j) {ans[j - n] = f[n][j] / m;}return ans;}
}

C++

class Solution {
public:vector<double> dicesProbability(int n) {int f[n + 1][6 * n + 1];memset(f, 0, sizeof f);for (int j = 1; j <= 6; ++j) {f[1][j] = 1;}for (int i = 2; i <= n; ++i) {for (int j = i; j <= 6 * i; ++j) {for (int k = 1; k <= 6; ++k) {if (j >= k) {f[i][j] += f[i - 1][j - k];}}}}vector<double> ans;double m = pow(6, n);for (int j = n; j <= 6 * n; ++j) {ans.push_back(f[n][j] / m);}return ans;}
};

Go

func dicesProbability(n int) (ans []float64) {f := make([][]int, n+1)for i := range f {f[i] = make([]int, 6*n+1)}for j := 1; j <= 6; j++ {f[1][j] = 1}for i := 2; i <= n; i++ {for j := i; j <= 6*i; j++ {for k := 1; k <= 6; k++ {if j >= k {f[i][j] += f[i-1][j-k]}}}}m := math.Pow(6, float64(n))for j := n; j <= 6*n; j++ {ans = append(ans, float64(f[n][j])/m)}return
}

JavaScript

/*** @param {number} n* @return {number[]}*/
var dicesProbability = function (n) {const f = Array.from({ length: n + 1 }, () => Array(6 * n + 1).fill(0));for (let j = 1; j <= 6; ++j) {f[1][j] = 1;}for (let i = 2; i <= n; ++i) {for (let j = i; j <= 6 * i; ++j) {for (let k = 1; k <= 6; ++k) {if (j >= k) {f[i][j] += f[i - 1][j - k];}}}}const ans = [];const m = Math.pow(6, n);for (let j = n; j <= 6 * n; ++j) {ans.push(f[n][j] / m);}return ans;
};

C#

public class Solution {public double[] DicesProbability(int n) {int[,] f = new int[n + 1, 6 * n + 1];for (int j = 1; j <= 6; ++j) {f[1, j] = 1;}for (int i = 2; i <= n; ++i) {for (int j = i; j <= 6 * i; ++j) {for (int k = 1; k <= 6; ++k) {if (j >= k) {f[i, j] += f[i - 1, j - k];}}}}double m = Math.Pow(6, n);double[] ans = new double[5 * n + 1];for (int j = n; j <= 6 * n; ++j) {ans[j - n] = f[n, j] / m;}return ans;}
}

Swift

class Solution {func dicesProbability(_ n: Int) -> [Double] {var f = Array(repeating: Array(repeating: 0, count: 6 * n + 1), count: n + 1)for j in 1...6 {f[1][j] = 1}if n > 1 {for i in 2...n {for j in i...(6 * i) {for k in 1...6 {if j >= k {f[i][j] += f[i - 1][j - k]}}}}}var m = 1.0for _ in 0..<n {m *= 6.0}var ans = Array(repeating: 0.0, count: 5 * n + 1)for j in n...(6 * n) {ans[j - n] = Double(f[n][j]) / m}return ans}
}

方法二：动态规划（空间优化）

我们可以发现，上述方法中的 $f[i][j]$ 的值仅与 $f[i-1][j-k]$ 有关，因此我们可以使用滚动数组的方式，将空间复杂度优化至 $O(6n)$。

Python3

class Solution:def dicesProbability(self, n: int) -> List[float]:f = [0] + [1] * 6for i in range(2, n + 1):g = [0] * (6 * i + 1)for j in range(i, 6 * i + 1):for k in range(1, 7):if 0 <= j - k < len(f):g[j] += f[j - k]  #仅仅与投掷n-1个骰子的方案数 数组有关f = gm = pow(6, n)return [f[j] / m for j in range(n, 6 * n + 1)]

Java

class Solution {public double[] dicesProbability(int n) {int[] f = new int[7];Arrays.fill(f, 1);f[0] = 0;for (int i = 2; i <= n; ++i) {int[] g = new int[6 * i + 1];for (int j = i; j <= 6 * i; ++j) {for (int k = 1; k <= 6; ++k) {if (j - k >= 0 && j - k < f.length) {g[j] += f[j - k];}}}f = g;}double m = Math.pow(6, n);double[] ans = new double[5 * n + 1];for (int j = n; j <= 6 * n; ++j) {ans[j - n] = f[j] / m;}return ans;}
}

Go

func dicesProbability(n int) (ans []float64) {f := make([]int, 7)for i := 1; i <= 6; i++ {f[i] = 1}for i := 2; i <= n; i++ {g := make([]int, 6*i+1)for j := i; j <= 6*i; j++ {for k := 1; k <= 6; k++ {if j-k >= 0 && j-k < len(f) {g[j] += f[j-k]}}}f = g}m := math.Pow(6, float64(n))for j := n; j <= 6*n; j++ {ans = append(ans, float64(f[j])/m)}return
}

JavaScript

/*** @param {number} num* @return {number[]}*/
var dicesProbability = function (n) {let f = Array(7).fill(1);f[0] = 0;for (let i = 2; i <= n; ++i) {let g = Array(6 * i + 1).fill(0);for (let j = i; j <= 6 * i; ++j) {for (let k = 1; k <= 6; ++k) {if (j - k >= 0 && j - k < f.length) {g[j] += f[j - k];}}}f = g;}const ans = [];const m = Math.pow(6, n);for (let j = n; j <= 6 * n; ++j) {ans.push(f[j] / m);}return ans;
};

C#

public class Solution {public double[] DicesProbability(int n) {int[] f = new int[7];for (int i = 1; i <= 6; ++i) {f[i] = 1;}f[0] = 0;for (int i = 2; i <= n; ++i) {int[] g = new int[6 * i + 1];for (int j = i; j <= 6 * i; ++j) {for (int k = 1; k <= 6; ++k) {if (j - k >= 0 && j - k < f.Length) {g[j] += f[j - k];}}}f = g;}double m = Math.Pow(6, n);double[] ans = new double[5 * n + 1];for (int j = n; j <= 6 * n; ++j) {ans[j - n] = f[j] / m;}return ans;}
}